PHP function not returning results to ajax call

Question

I went through solutions provided to similar questions in Stack Overflow and other as well. (Seems like its not only me with this problem ). So I finally am posting the question with a hope to find some solution.

I am sending two numbers to PHP function via ajax call and expect to get sum from function. I think I am right not sure though, that half of the ajax is right and half not i.e. I can send values to PHP function (i think) but I can't get sum instead in success message I get "html data of the page" :D

Can you please point my mistake and viable solution.

Ajax:

$(document).ready(function(){
    $('#myForm').submit(function(event){
    
        var num1 = Number($('#num1').val());
        var num2 = Number($('#num2').val());
        var summ = num1+num2;
        //var data = $('#myForm').serialize();
        console.log(" nums are " + summ); //totally works
        var myNums = "num1= " + num1 + " & num2=" + num2; 
        alert(myNums);       //totally works
        event.preventDefault();
        $.ajax({
            url:'arrayTest.php',
            type: 'POST',
            data: {num1 : num1,
                    num2 : num2 },
            success:function(data){
                console.log(" <br>success yr data is " + myNums);
                var results = data;
                alert(data); // i get html of the entire page containing form
                return false;
            },
            error:function(error){
                console.log(" sorry error " + error); // get this when i add header in php 
            }
        });
    });

});

html :

<form id ="myForm" method="post" action="">  

  number1: <input type="number" name="num1" id="num1">
  <br><br>
  number2: <input type="number" name="num2" id="num2">
  <br><br>
   <br><br>
  <input type="submit" id ="submit" name="submit" value="add">  
</form>

php:

if(isset($_POST['submit'])){
            
    $num1 = $_POST['num1'];
    $num2 = $_POST['num2'];
    $result= $num1 + $num2;
    //echo "num1 is ".$num1;// **i didnt see any echo** 
    
    echo " your result ". $result; // **i didnt see any echo** 
    return ($result);
}

PS: html and ajax is in same page, php function in another; I see message

XHR finished loading: POST "http://localhost/phpTest/arrayTest.php"."

in console (I don't have idea what it means!)

If I'm not wrong, your check for $_POST['submit'] is the culprit here. The POST data you send along with your request doesn't have a key called 'submit' in it, so the PHP code never executes like you expect. — Connor
– Connor, Commented Oct 12, 2018 at 20:58
I strongly suggest you utilize the browser's Dev Tools. You can examine exactly what gets sent and received from an AJAX query - it takes a lot of the guesswork out. — Quasipickle
– Quasipickle, Commented Oct 12, 2018 at 21:05
@Connor Thank you for your suggestion. Actually if i am not wrong you meant the i should have used $_POST['add'] or something? If yes, sorry to tell that it had not helped me actually .... — vins
– vins, Commented Oct 13, 2018 at 5:58

Ochui Princewill · Accepted Answer · 2018-10-12 21:42:12Z

1

Your form define num1, num2 and add but you are using submit which is not define.

if(isset($_POST['add'])){

  $num1 = $_POST['num1'];
  $num2 = $_POST['num2'];
  $result= $num1 + $num2;
  //echo "num1 is ".$num1;// **i didnt see any echo** 

   echo " your result ". $result; // **i didnt see any echo** 
   return ($result);
}

TIPS: Always turn on error reporting and development Environment and use print_r() to inspect variable

answered Oct 12, 2018 at 21:42

Ochui Princewill

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5 Comments

vins Over a year ago

Thank you friend. Turns out that i tried your solution beforehand i posted here ..I thought it could be one of reason so I had been trying to change 'POST' to 'add' and vice-versa.

Ochui Princewill Over a year ago

UPDATE: use print_r($_POST) and notice sudmit/add was sent with the form

vins Over a year ago

..i used it in php page and no error found (unless i used it before the function/if(isset...))

Ochui Princewill Over a year ago

sorry for the typo i meant add/submit is'n part of the post data. Perhaps you should consider adding it with JavaScript or replace isset($_POST['add'])** with **isset($_POST)

vins Over a year ago

Princewell Thanks a lott my friend -Your solution is perfect ..I now got print_r($num1+$num2) in php file as alert ...but still this is not flashing in ajax "data/result"

Stefan Janjic · Accepted Answer · 2018-10-12 22:24:30Z

0

If your code works just disable XMLHttpRequests from chrome console.

open the console -> click the vertical "..." in the top right -> go to settings, and uncheck "Log XMLHttpRequests" under the "Console" header.

answered Oct 12, 2018 at 22:24

Stefan Janjic

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